Question

Consider a causal LTI system implemented as the RL circuit shown below. In this circuit, v(t) is the input voltage. The current i(t) is considered the system output. i(t) R L wwwm v(t) (a) Find the differential equation relating v(t) and i(t). (b) Determine the frequency response of this system (H(jw)). (c) Determine the output it) if v(t) = sin(t), R=10 and L=1. (d) Sketch Bode plot of H (jw) for R=10 and L=1. (e) Determine if the system is a lowpass filter, highpass filter, or none.

Answer 1

Solution Güven circuit mm ults = input VCL) ilts = output to kvL in the ciruit from - -u(t) ti (1) R+ L dilts to dt voltage across o Inductor L dilt) tilor = VCt) - T dict) +i (t) R/ L = V(t) ] viju SL o frequence response H (gw) = output input = (to Iliw) =) Write the circuit elements in laplace R un mom 1.) V(6 17 (5) R I I) + SLI()=0 ho I 10 Rish] = v(s) . - 19 -1 KUL (s) = RISL - L RISL

3 H(S) - 10 - at Substitute s=jw Hljus 1(jw) - On vw) 1 JWLtr

a ut- tint =) W- 1 rad see R = lor, i=1 H = JXL = 1 WL=i1 R JXL ilus on .) a les ucts = 120 vt) DE ILO - ilm = - VCE) RtjXL Ilo lotj 1 = 1 ko Vroot 12 Ltant (1) 10.049 C5.710 = 0.0995 L-5.710 = 0.01L-5.710 - (0.01 sin (t-5.710) =ilt)) (OR) - VC VLES (jw) Wahoo toto from Sinusoidal input Ult) = A sinut+o) » i(t) = A xH (jw) Sinfurt gaten (H1) (wewe in the given problem Wot 1, vcts = sint isty = A (HGjwlown sin(t +e)

Vw.)?7R² V 1²+ 102 10.oug 0.0995 - e 0.01 * BI LH (gw) 0= - tant Ecoo Wz cod = -tan (1) Os -5.710 3) i(t) = 1.0.01). Sin It-05.710) i(t)= 0.01 sin (t - 5.10)+) H6w) = - JOLER -) H(S) = 1 SLAR SLAR - (SL) HOR +1 here Cooner frequency we = (B/ To = 10 rassec

I since there is no pole at origin before 10 gad/sec, the plot will be a straight line parallel to w-axis. - And then the plot has negative slope after lorad/sec, since it isa pole of order && degree=1 - the bode plot » l lH;ros l et ( filo? V ht" for approximate - in dB 3 20log. H(jw) on atyplot Rolog wie » - Bologie de (Cias de » CHIju) c -tan [ ww] = -tan (6) - Cets see the vartiation of magnitude & phase with change in w w/ Hj w l de We 10 =) WLLWC 2) W CC 120 | 3) - 20log 10+1=0 dB Ode atw=10 -)-20 log 10 = odb ods

at We 100 -200g 10010 -20dB doo wa 1000 - 2000g 1000= 10 -40dB Frode I w WKL Wc I LH (jw) s want - - tań (07-01 w = 10 > tan" ( 1 = 45° Lontoo Is tan for wzowe wypu -) tan (w] = 90 s [Helde -20 dB/dec 10 100 1000 - 2008 - podo ALHw) yo. Asymptotic pole nexactplot

) magnita de plot of HjW) resembles the property of a high pass filter where frequencies below certain frequency Called Comer frequency or cut off frequency the signal is o iey attenuated by above that frequency signal is passed